What volume of nitrogen gas is produced by the explosion of 65.0 g of sodium azide at stp? the balanced chemical equation is 2 nan3(s) = 2 na(s) + 3 n2(g) for this reaction.

a. 33.6 l
b. 67.2 l
c. no right choice.

a) V = 33.6 L

Explanation:

Given data:

Mass of sodium azide = 65 g

Volume of Nitrogen = ?

Solution:

Chemical equation:

2NaN₃  → 3N₂ +2Na

Number of moles of NaN₃:

Number of moles of NaN₃ = Mass /molar mass

Number of moles of NaN₃ = 65 g / 65 g/mol

Number of moles of NaN₃ = 1 mol

Now we will compare the moles of NaN₃ with N₂ .

                  NaN₃        :              N₂

                   2             :               3

                   1              :            3/2 = 1.5 mol

Volume of nitrogen gas:

PV = nRT

V = nRT /P

V = 1.5 mol. 0.0821 atm.L. mol⁻¹ .K⁻¹ × 273.15 K / 1 atm

V = 33.64 L